The correct graph between the frequency n and square root of density `rho` of a wire, keeping its length, radius and tension constant, is

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Explanation

To determine the correct graph between the frequency \( n \) and the square root of density \( \rho \) of a wire, while keeping its length, radius, and tension constant, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Relationship**:
The fundamental frequency \( f \) of a wire can be expressed in terms of tension \( T \), length \( L \), and linear mass density \( \mu \):
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]

2. **Express Linear Mass Density**:
The linear mass density \( \mu \) can be expressed as:
\[ \mu = \frac{m}{V} = \frac{m}{A \cdot L} = \frac{\rho \cdot A}{L} \]
where \( A \) is the cross-sectional area and \( \rho \) is the density.

3. **Substitute for Linear Mass Density**:
Substitute \( \mu \) back into the frequency equation:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\rho \cdot A}} \]

4. **Identify Constants**:
Since \( L \), \( A \), and \( T \) are constant, we can define a new constant \( k \):
\[ k = \frac{1}{2L} \sqrt{\frac{T}{A}} \]
Thus, the frequency can be rewritten as:
\[ f = k \cdot \frac{1}{\sqrt{\rho}} \]

5. **Relate Frequency to Density**:
This implies that:
\[ n \propto \frac{1}{\sqrt{\rho}} \]
or
\[ n = \frac{k}{\sqrt{\rho}} \]

6. **Graphical Representation**:
If we let \( x = \sqrt{\rho} \), then:
\[ n = \frac{k}{x} \]
This represents a hyperbolic relationship between \( n \) and \( \sqrt{\rho} \).

7. **Conclusion**:
The graph of \( n \) versus \( \sqrt{\rho} \) will be a hyperbola, indicating that as the square root of density increases, the frequency decreases.